Answer:
Consider f: N → N defined by f(0)=0 and f(n)=n-1 for all n>0.
Step-by-step explanation:
First we will prove that f is surjective. Let y∈N be any natural number. Define x as the number x=y+1. Then x∈N, and f(x)=x-1=(y+1)-1=y. We conclude that f is surjective.
However, f is not injective. Take x1=0 and x2=1. Then x1≠x2 but f(x1)=0 and f(x2)=x2-1=1-1=0. We have shown that there are two natural numbers x1,x2 such that x1≠x2 but f(x1)=f(x2), that is, f is not injective.
Note:
If 0∉N in your definition of natural numbers, the same reasoning works with the function f: N → N defined by f(1)=1 and f(n)=n-1 for all n>1. The only difference is that you consider x1=1, x2=2 for the injectivity.
The way we simplify this is by using the distributive property and multiply every term by 0.5. When we do this we get 2a+3b
5 = what percent of 23
5 = x% of 23
5 = (x/100)*23
5 = 23x/100
5*100 = 23x
23x = 5*100
x = 5*100/23
x = 21.739
5 is ≈ 21.739% of 23
Answer:
Slope: −1
y-intercept: (0,3)
Step-by-step explanation:
